Question

What is the percent-by-mass concentration of a 100.0-mL \(\mathrm{H}_{3} \mathrm{PO}_{4}\) solution that contains \(88.20 \mathrm{~g}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ? The density of the solution is \(1.40 \mathrm{~g} / \mathrm{mL}\).

Short answer

The percent-by-mass concentration of the \(H3PO4\) solution is 63.0%.

Step-by-step solution

Calculate the Mass of the Solution

Firstly, the mass of the solution needs to be calculated. The total mass of the solution can be determined by multiplying the volume of the solution by the density. Given a volume of \(100.0 mL\) and a density of \(1.40 g/mL\), we can calculate the mass by using the formula \(mass = density \times volume\). So the total mass of the solution is \( \)100.0 mL* \(1.40 g/mL = 140.0 g\).

Calculate the Concentration of Solution

The percent-by-mass concentration of a solution is given by the formula: \(\ percent-by-mass = \frac {mass \ of \ solute}{mass \ of \ solution} \times 100\%. \) Given that the mass of \(H3PO4\) is \(88.20g\) and we just calculated the total mass of the solution as \(140.0 \ g\), we can substitute these values into our formula.

Final Calculation

Substituting the calculated values into the formula gives us: \(\ percent-by-mass = \frac {88.20 \ g}{140.0 \ g} \times 100% = \) 63.0%. Therefore, the percent by mass of \( H3PO4\) in the solution is 63.0%.

Key concepts

H3PO4 solution

An \( \mathrm{H}_{3}\mathrm{PO}_{4} \) solution refers to a mixture in which phosphoric acid (\( \mathrm{H}_{3}\mathrm{PO}_{4} \)) is dissolved in a solvent, typically water. Phosphoric acid is a common inorganic acid used in various industrial and laboratory applications. It is a colorless, odorless, and non-volatile liquid. In solution, it behaves as a weak acid, meaning it partially ionizes in water.

• Properties of \( \mathrm{H}_{3}\mathrm{PO}_{4} \): It is known for being non-toxic, which makes it safe for certain consumer products.
• Applications: It is used in fertilizers, cleaning products, food additives, and as a buffering agent in certain chemical reactions.

In any solution, including \( \mathrm{H}_{3}\mathrm{PO}_{4} \), the solute (such as \( \mathrm{H}_{3}\mathrm{PO}_{4} \)) is the substance dissolved in the solvent, and it determines the chemical concentration of the solution.

density and volume

Density is a measure of how much mass exists in a given volume. For any liquid solution, knowing its density and volume is crucial. Density is usually expressed in units of grams per milliliter (g/mL) or kilograms per liter (kg/L). The formula to find the density (\( \rho \)) of a substance is given by:\[ \rho = \frac{\text{mass}}{\text{volume}} \]This formula can be manipulated to find mass or volume if the other variables are known.

• Effect of Temperature: Remember, the density of a solution can change with temperature, as the volume of liquids usually expands or contracts due to heat changes.

In our specific example, the density of the \( \mathrm{H}_{3}\mathrm{PO}_{4} \) solution was provided as \(1.40 \mathrm{~g}/\mathrm{mL}\). For a volume of \(100.0 \mathrm{~mL}\), the total mass of the solution was found by multiplying density by volume, resulting in \(140.0 \mathrm{~g}\). Understanding this relationship is fundamental when calculating concentrations.

chemical concentration calculations

Chemical concentration refers to the amount of solute present in a given quantity of solvent or solution. It is typically expressed in forms like molarity, molality, volume percent, and as in our case, percent-by-mass. Here, we deal specifically with percent-by-mass concentration which is calculated using the formula:\[\text{percent-by-mass} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100\%\]This formula tells us what fraction of the total solution’s mass is due to the solute. It's helpful in situations where precise chemical composition needs to be known, for example, when preparing solutions for reactions.

• Importance: Knowing the concentration is essential for ensuring consistency in experiments and productions, as it affects reaction rates and outcomes.

For the exercise, with \(88.20 \mathrm{~g}\) of \( \mathrm{H}_{3}\mathrm{PO}_{4} \) in a \(140.0 \mathrm{~g}\) solution, the percent-by-mass was calculated to be \(63.0\%\). This measure provides insight into how much of the solution is made up of \( \mathrm{H}_{3}\mathrm{PO}_{4} \).